MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 12

Given

$a_3=15, S_{10}=125$ , Find d and

$a_{10}$

0%
$18$
0%
$4$
0%
$8$
0%
$6$

Explanation

As we know that the nth term,

$a_n = a+(n-1)d$

& Sum of first

$n$ terms,

$S_n = \dfrac{n}{2} (2a+(n-1)d)$ , where

$a$ &

$d$ are the first term & common difference of an AP.

$S_{10}=125=\dfrac {10}{2}(2a+9d)$

$\Rightarrow 2a+9d=25$ ............... (i)

$a_3=a+2d=15$ ............. (ii)

Subtracting equation (ii) from (i) we get

$a+7d=10$ (the

$8^{th}$ term)

Subtracting

$3^{rd}$ term from

$8^{th}$ term we get

$(a+7d)-(a+2d)=10-15=-5$

$\Rightarrow 5d=-5$

$\Rightarrow d=-1$

So,

$a=17$

$a_{10}=17-9=8$

Find the sum of the first

$15$ positive odd numbers.

0%
$225$
0%
$200$
0%
$210$
0%
$217$

Explanation

First $15$ positive odd numbers will be,

$1,3,5,7,9,\dots$

We can observe that these numbers are in A.P.

The sum of first $n$ terms of an A.P whose first term is $a$ and common difference is $d$ can be written as, $S_n = \dfrac{n}{2} [2a + (n - 1)d].$

From the series of first positive odd numbers we can conclude that, $a=1$ and $d=2.$

Substitute the values of $a,\ d$ and $n$ in the formula of sum of first $n$ terms of an A.P,

$\begin{aligned}{}{S_{15}} &= \frac{{15}}{2}[2 \times 1 + (15 - 1)2]\\& = 7.5[2 + 14\times2]\\& = 7.5[2 + 28]\\ &= 7.5[30]\\& = 225\end{aligned}$

So, the sum of the first $15$ positive odd numbers of the A.P. is $225.$

Find the sum of first 32 terms of the arithmetic series if

$a_1 = 12$ and

$a_{32} = 40$ .

0%
$830$
0%
$831$
0%
$832$
0%
$834$

Explanation

Given:
First term

$= 12$
last term

$= 40$

The sum of

$n$ terms of arithmetic sequence can be calculated by the formula

$S_n = \dfrac{n}{2} (a_1 + a_n)$
Where

$n$ is the number of terms,

$a_1$ is the first term and

$a_n$ is the last term.

Here,

$n = 32$

$\therefore\ \ S_{32} = \dfrac{32}{2} (12 + 40)$

$= 16 (12 + 40)$

$= 16 (52)$

$= 832$

Hence, the sum of the first

$32$ terms is

$832$

Given

$d=5, S_9=75$ , find a

0%
$\dfrac{-35}{3}$
0%
$\dfrac{-28}{3}$
0%
$\dfrac{-54}{3}$
0%
$\dfrac{-34}{3}$

Explanation

We know that sum of first

$n$ terms,

$S_n = \dfrac{n}{2} (2a+(n-1)d)$ , where

$a$ &

$d$ are the first term & common difference of an AP.

$\therefore S_9=\dfrac {9}{2}(2a+d(9-1))$

$\Rightarrow 75=4.5(2a+8d)$

$\Rightarrow 2a+8d=\dfrac {75}{4.5}$

$\Rightarrow 2a=\dfrac {150}{9}-40=\dfrac {150-360}{9}=\dfrac {-210}{9}=\dfrac {-70}{3}$

$\Rightarrow a=\dfrac {-35}{3}$

Calculate the sum of whole numbers between 0 and 79 which are divisible by 11.

0%
$300$
0%
$310$
0%
$308$
0%
$320$

Explanation

The numbers between

$0$ and

$79$ which are divisible by

$11$ are

$11, 22, 33, 44, ...... 77$ .

The sum of first

$n$ terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)

Where

$n =$ number of terms = ?

$a =$ first odd number

$=11$

$d =$ common difference of A.P.

$=11$

$T_n =$ Last term

$=77$

$T_n = a + (n - 1)d$

$77 = 11 + (n - 1) 11$

$77 = 11 + 11n - 11$

$77 - 11 + 11 = 11n$

$77 = 11n$

$\Rightarrow n = 7$

Apply the given data in the formula

$(1)$ , we get

$S_{7} = \dfrac{7}{2} [2 \times 11 + (7 - 1)11]$

$= 3.5 [22 + (6)11]$

$= 3.5 [22 + 66]$

$= 3.5 [88]$

$= 308$

So, the sum of whole numbers between

$0$ and

$79$ which are divisible by

$11$ is

$308$ .

Find the sum of first 20 even terms of the arithmetic sequence.

0%
$420$
0%
$401$
0%
$402$
0%
$403$

Explanation

The even series will be

$2,4,6,8,10......40$

The sum of n terms of an arithmetic sequence can be calculated by

$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$

Where,

$n =$ number of terms

$\Rightarrow$ 20

$a =$ first even term

$\Rightarrow$ 2

$d =$ common difference of A.P.

$\Rightarrow$ 2

substitute the given data in the formula,

$\Rightarrow S_{20} = \dfrac{20}{2} [2 \times 2 +(20 - 1)2]$

$= 10[4 + 38]$

$= 10 [42]$

$= 420$

$\therefore$ the sum of first

$20$ even terms of the A.P. is

$420$ .

Find the sum of first 10 terms of the arithmetic series if

$a_1 = 25$ and

$a_{10} = 100$ .

0%
$300$
0%
$320$
0%
$625$
0%
$312$

Explanation

The sum of n terms of arithmetic sequence can be calculated by the formula

$S_n = \dfrac{n}{2} (a_1 + a_n)$
Where is the number of terms,

$a_1$ is the firth term and

$a_n$ is the last term.

$n = 50$
First term

$= 25$
Tenth term

$= 100$
Therefore,

$S_{10} = \dfrac{10}{2} (25 + 100)$

$= 5 (25 + 100)$

$= 5 (125)$

$= 625$
Hence, the sum of first 10 terms is

$625.$

In an Arithmetic progression, the first term

$a = 10$ , and last term,

$l = 100$ . The number of terms is

$1000$ . Find their sum.

0%
$52,000$
0%
$53,000$
0%
$54,000$
0%
$55,000$

Explanation

Given,

$a=10, \text{last term }=100$ and

$\text{number of terms}=1000$

Sum of first

$n$ terms in an A.P. is

$S_n = \dfrac{n}{2}$ [First term

$+$ Last term]

$= \dfrac{1000}{2} [10 + 100]$

$= 500 [110]$

$S_{1000} = 55,000$

Find the sum of even numbers between

$11$ and

$51$ .

0%
$600$
0%
$610$
0%
$620$
0%
$630$

Explanation

The sum of first n terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
The list of even numbers between

$11$ and

$51$ is

$12, 14, 16, 18,....... 50$
Where

$n$ = number of terms
a = first even number

$\Rightarrow$ 12
d = common difference of A.P.

$\Rightarrow$ 2

$T_n =$ Last term

$\Rightarrow$ 50

$T_n = a + (n - 1)d$

$50 = 12 + (n - 1) 2$

$50 = 12 + 2n - 2$

$50 - 12 + 2 = 2n$

$40 = 2n$

$n = 20$
Apply the given data in the formula (1),

$S_{20} = \dfrac{20}{2} [2 \times 12 + (20 - 1)2]$

$= 10 [24 + (19)2]$

$= 10 [24 + 38]$

$= 10 [62]$

$= 620$
So, the sum of even numbers between

$11$ and

$51$ is

$620$ .

Find the sum of fterms of AP whose first term is 26 and last term is 14.

0%
200
0%
250
0%
300
0%
400

Explanation

The sum of first n terms of arithmetic series formula can be written as,

$S_{n}=\dfrac{n}{2}$ [First term + Last term] ... eq ( 1 )

n = number of terms = 20

First term = 26

Last term = 14

$\therefore S_{n}=\dfrac{20}{2} [26+14]=400$ (Ans

$\rightarrow$ D)

Find the sum of odd numbers between

$0$ and

$90$ .

0%
$2025$
0%
$2020$
0%
$2035$
0%
$2041$

Explanation

The sum of first n terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
The list of odd numbers between

$0$ and

$90$ is

$1, 3, 5, 7, 9, .......... 89$
Where

$n =$ number of terms

$a =$ first odd number

$\Rightarrow$

$1$

$d =$ common difference of A.P.

$\Rightarrow 2$

$T_n =$ Last term

$\Rightarrow$ 89

$T_n = a + (n - 1)d$

$89 = 1 + (n - 1) 2$

$89 = 1 + 2n - 2$

$89 - 1 + 2 = 2n$

$90 = 2n$

$n = 45$
Apply the given data in the formula (1),

$S_{45} = \dfrac{45}{2} [2 \times 1 + (45 - 1)2]$

$= 22.5 [2 + (44)2]$

$= 22.5 [2 + 88]$

$= 22.5 [90]$

$= 2025$
So, the sum of odd numbers between

$0$ and

$90$ is

$2,025$ .

What is the sum of first

$100$ multiples of

$12$ ?

0%
$60600$
0%
$61200$
0%
$62900$
0%
$60500$

Explanation

Frist

$100$ multiples of

$12$ will be,

$12,24,36,\dots$

So, we can conclude that the multiples of

$12$ form an arithmetic progression whose first term is

$12$ and the common difference is

$12.$

The sum of first

$n$ terms of an arithmetic series whose first term is

$a$ and common difference is

$d$ can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$

So, by using above formula,

$S_{100} = \dfrac{100}{2} [2 \times 12 + (100 - 1)12]$

$= 50 [24 + (99)12]$

$= 50 [24 + 1188]$

$= 50 \times1212$

$= 60,600$

So, the sum of the first

$100$ multiples of

$12$ is

$60600.$

Calculate the sum of even numbers between

$12$ and

$90$ which are divisible by

$8$ .

0%
$500$
0%
$510$
0%
$520$
0%
$620$

Explanation

The list of odd numbers between

$12$ and

$90$ which are divisible by

$8$ are

$16, 24, ..... 88$

The sum of first

$n$ terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)

Where

$n =$ number of terms = ?

$a =$ first odd number

$=16$

$d =$ common difference of A.P.

$=8$

$T_n =$ Last term

$=88$

$T_n = a + (n - 1)d$

$88 = 16 + (n - 1) 8$

$88 = 16 + 8n - 8$

$88 - 16 + 8 = 8n$

$80 = 8n$

$n = 10$

Apply the given data in the formula (1),

$S_{10} = \dfrac{10}{2} [2 \times 16 + (10 - 1)8]$

$= 5 [32 + (9)8]$

$= 5 [32 + 72]$

$= 5 [104]$

$= 520$

So, the sum of even numbers between

$12$ and

$90$ which are divisible by

$8$ is

$520$ .

Calculate the sum of even numbers between 100 and 150 which are divisible by 13.

0%
$494$
0%
$410$
0%
$420$
0%
$430$

Explanation

The even numbers between

$100$ and

$150$ which are divisible by

$13$ are

$104, 117, ............ 143$

The sum of first

$n$ terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)

Where

$n =$ number of terms = ?

$a =$ first odd number

$=104$

$d =$ common difference of A.P.

$=13$

$T_n =$ Last term

$=143$

$T_n = a + (n - 1)d$

$143 = 104 + (n - 1) 13$

$143 = 104 + 13n - 13$

$143 - 104 + 13 = 13n$

$52 = 13n$

$n = 4$

Apply the given data in the formula (1),

$S_{4} = \dfrac{4}{2} [2 \times 104 + (4 - 1)13]$

$= 2 [208 + (3)13]$

$= 2 [208 + 39]$

$= 2 [247]$

$= 494$

So, the sum of even numbers between

$100$ and

$150$ which are divisible by

$13$ is

$494$ .

Find the sum of even numbers between 1 and 25.

0%
$155$
0%
$156$
0%
$157$
0%
$158$

Explanation

The sum of first n terms of arithmetic series formula can be written as,

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$ ............ (1)
The list of even numbers between 1 and 25 is 2, 4, 6, 7, 8, .......... 24
Where n = number of terms = ?
a = first even number

$a = 2$
d = common difference of A.P.

$d=2$

$T_n =$ Last term

$T_n = 24$

$T_n = a + (n - 1)d$

$24 = 2 + (n - 1) 2$

$24 = 2 + 2n - 2$

$24 - 2 + 2 = 2n$

$24 = 2n$

$n = 12$
Apply the given data in the formula (1),

$S_{12} = \dfrac{12}{2} [2 \times 2 + (12 - 1)2]$

$= 6 [4 + (11)2]$

$= 6 [4 + 2]$

$= 6 [26]$

$= 156$
So, the sum of even numbers between 1 and 25 is 156.

$a_3 = 9$ and

$a_7 = 21$ . Find

$S_5$

0%
40
0%
45
0%
50
0%
55

Explanation

Using the formula,

$a_n = a + (n - 1) d$
So,

$a_3 = a + (3 - 1) d$

$a = a + 2d$ ......... (1)

$a_7 = a + (7 - 1) d$

$21 = a + 6d$ ........ (2)
Subtracting equation (1) & (2)

$a = a + 2d$ ........ (1)

$21 = a = bd$ ....... (2)
(-) (-) (-)
------------------

$- 20 = -4d$

$d = 5$
Put the value of d in equation (1)

$a = a + 2(5)$

$a = - 1$
The sum of first n term is given by

$S_n = \dfrac{n}{2} [2a + (n - 1)d]$

$S_5 = \dfrac{5}{2} [2(-1) + (5 - 1) 5]$

$= \dfrac{5}{2} (-2 + 20)$

$= 2.5 (18)$

$S_5 = 45$

Check whether

$80$ is a term of the AP.

$0, 0.5, 1, 1.5,\dots$

0%
True
0%
False
0%
Data Insufficient
0%
Can't Say

Explanation

$0, 0.5,1,1.5,\dots$

Here,

$a = 0, d = 0.5$
Let

$80$ be the

$n^{th}$ term of this AP.

Since,

$a_n = a + (n - 1) d$

$\Rightarrow 80 = 0 + (n - 1) 0.5$

$\Rightarrow 80 = 0.5 n - 0.5$

$\Rightarrow 80 + 0.5 = 0.5 n$

$\Rightarrow n = 161$

Clearly,

$n$ is an integer.
Therefore,

$80$ is the

$161^{th}$ term of this AP.

Find the

$12^{th}$ term from the end of the A.P.

$2, 6, 10 ......... 58$

0%
$22$
0%
$14$
0%
$16$
0%
$28$

Explanation

The given A.P. can be written in reverse order as

$58 .......... 10, 6, 2$

Here

$a = 58, d = -4, n = 12$

$a_{12} = a + (12- 1) d$

$= 58 + 11 \times (-4)$

$= 58 - 44$

$a_{12} = 14$

$\therefore$

$14$ is the

$12^{th}$ term from the end.

Find the common difference for the arithmetic sequence whose formula is

$a_n = 12n - 7$

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

Assume

$n = 1, 2, 3,$ .......
Put those values in

$a_n = 12n - 7$

$a_1 = 12 (1) - 7 = 5$

$a_2 = 12 (2) - 7 = 17$

$a_3 = 12 (3) - 7 = 29$

$a_4 = 12 (4) - 7 = 41$
Therefore, the common difference,

$d$ is

$12$

Find the common difference, if

$a_2 = 10$ and

$a_5 = 20$ .

0%
$3.33..........$
0%
$3.35..........$
0%
$3.65..........$
0%
$3.75..........$

Explanation

The nth term of A.P. is

$a_n = a_1 + (n - 1) d$

$\therefore a_2 = a + (n -1) d$

$10 = a + (2- 1) d$

$10 = a + d$ ..... (1)

$a_5 = a + (n - 1) d$

$20 = a + 4d$ ..... (2)
Equating equation (1) & (2)

$10 = a + d$

$20 = a + 4d$
(-) (-) (-)
-----------------

$-10 = - 3d$

$d = \dfrac{+10}{+3}$

$d = 3.33 ........$

The arithmetic sequence is given as:- -7, -4, -1, 2, 5 .......... Find their common difference and next term of the sequence.

0%
Common difference $= 3$ ; $a_6 = 8$
0%
Common difference $= 2$ ; $a_4 = 7$
0%
Common difference $= 1$ ; $a_6 = 5$
0%
Common difference $= 0$ ; $a_6 = 0$

Explanation

$a_1 = - 7$

$a_2 = - 4$
Common difference,

$d = a_2 - a_1$

$= -4 - (- 7)$

$= -4 + 7 = 3$

$d = 3$
The next term will be

$a_6$ So,

$a_5 + d = a_6$

$5 + 3 = 8$

$\therefore$ The next term is 8

$a_5 = 12$ and

$a_{15} = 400$ . Find

$S_{10}$

0%
314
0%
320
0%
316
0%
318

Explanation

Using the formula,

$a_n = a + (n - 1) d$
So,

$a_5 = a + (5 - 1) d$

$12 = a + 4d$ ............. (1)

$a_{15} = a + (15 - 1) d$

$400 = a + 14 d$ .......... (2)
Solving equation (1) & (2)

$12 = a + 4 d$

$400 = a + 14 d$
(-) (-) (-)
---------------------

$- 388 = -10d$

$d= 38.8$
Put the value of d in equation (1)

$12 = a + 4 (38.8)$

$12 = a + 155. 2$

$a = - 143.2$
The sum of first n term is given by

$S_n = \dfrac{n}{2} (2a + (n - 1)d)$

$S_{10} = \dfrac{10}{2} (2 (-143. 2) + (10 - 1) 38.8)$

$= 5 (- 286.4 + 349.2)$

$= 5 (62.8)$

$S_{10} = 314$

In an arithmetic series, find the sum of the first 10 terms if the first term is 3 and the common difference is 4.

0%
$110$
0%
$210$
0%
$310$
0%
$410$

Explanation

First term = a = 3
n = 10
d = 4

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

$S_{10}=\frac{10}{2}[2\times 3+(10-1)4]$

$S_{10}=5[6+36]$

$S_{10}=210$

In an arithmetic sequence

$1, -2, -5, -8$ ....., then find the

$12^{th}$ term.

0%
$-30$
0%
$-32$
0%
$33$
0%
$34$

Explanation

$12^{th}$ term of the A.P. is

$a_{12} = a + (n - 1) d$
Given series is

$1,-2,-5,-8...$

Here

$a = 1, d = - 3, a_{12} = ?$

Since

$a_{n} = a + (n-1)(d)$
Therefore,

$a_{12} = 1 + (12- 1) (- 3)$

$a_{12} = 1 + 11 \times (-3)$

$a_{12} = 1 - 33$

$\Rightarrow a_{12} = - 32$

Find the common difference in the sequence:

$7, 14, 21, ............ 70$

0%
$4$
0%
$5$
0%
$6$
0%
$7$

Explanation

First term,

$a_1 = 7$
Second term,

$a_2 = 14$
Common difference,

$d = a_2 - a_1$

$= 14 - 7 \Rightarrow 7$

$\therefore$ The common difference is 7.

What is the common difference, if

$a_1 = 100$ and

$a_2 = 250$

0%
$100$
0%
$150$
0%
$200$
0%
$210$

Explanation

Common difference,

$d = a_2 - a_1$

$= 250 - 100$

$d = 150$

A bus can travel

$10$ m in the first minute,

$25$ m the next minute,

$40$ m the third minute and so on in an arithmetic sequence. What is the total distance the bus travels in

$10$ minutes?

0%
$2,000$
0%
$775$
0%
$4,000$
0%
$5,000$

Explanation

The sequence is 10, 25, 40 .....

$a = 10, d = 15$
First find the common difference:

$a_n = a_1 + (n - 1) d$

$a_{10} = 10 + (10 - 1) 15$

$= 10 + 9 \times 15$

$a_{10} = 145$
We know that,

$S_n = \dfrac{n}{2}$ [First term + Last term]

$= \dfrac{10}{2} [10 + 145]$

$= 5 [155]$

$S_{10} = 775$
Hence, the bus will travel 775 m in 10 minutes.

10 people live on the 1st floor of the apartment, 22 people on the second floor and 34 people on the third floor, and so on in an arithmetic sequence up to the 10th floor. What is the total number of people living in the apartment?

0%
656
0%
564
0%
640
0%
589

Explanation

The sequence is 10, 22, 34 ......

$a = 10, d = 22-10 = 12$

$a_n = a_1 + (n - 1) d$

$a_{10} = 10 + (10 - 1) 12$

$= 10 + 9 \times 12$

$a_{10} = 118$

We know that,

$S_n = \dfrac{n}{2}$ [First term + Last term]

$= \dfrac{10}{2} [10 + 118]$

$= 5 [128]$

$S_{10} = 640$

Hence, the total number of people living in the apartment is

$640$ .

There are 25 tourist in the first travel, 50 tourist the second travel and 75 tourist the third travel in the museum and so on in an arithmetic sequence. Find the total number of tourist in 25 travels?

0%
8,460
0%
8,450
0%
8250
0%
8,125

Explanation

The sequence is 25, 50, 75

$a = 25, d = 25$
First find the common difference:

$a_n = a_1 + (n - 1) d$

$a_{25} = 25 + (25 - 1) 25$

$= 25 + 24 \times 25$

$a_{25} = 625$
We know that,

$S_n = \dfrac{n}{2}$ [First term + Last term]

$= \dfrac{25}{2} [25 + 625]$

$= 12.5 [650]$

$S_{25} = 8,125$
So, the total number of tourist in 25 travels is 8,125

What is the

$21^{st}$ term of the arithmetic sequence

$-5, -2, 1, 4, 7 ......?$

0%
$51$
0%
$55$
0%
$54$
0%
$53$

Explanation

Given sequence is

$-5, -2, 1, 4, 7 .........$
To find out: The

$21^{st}$ term of the sequence.

We know that, the

$n^{th}$ term of an A.P. is

$t_n=a+(n-1)d$

Here

$a = - 5,\ n=21\ and\ d = 3$

$\therefore \ t_{21} = - 5 + (21 - 1) 3$

$\Rightarrow - 5 + (20 ) 3$

$\Rightarrow - 5 + 60$

$\therefore \ t_{21} = 55$

Hence, the

$21^{st}$ term of the given sequence is

$55$ .

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 12 - MCQExams.com

0:0:8

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 12 - MCQExams.com

0:0:8

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.